Chapter 8 - Section 8.2 - Multiplying and Dividing Radicals - Exercise Set - Page 585: 104

$2\sqrt[5]{2}$

Using $\sqrt{x}\cdot\sqrt{y}=\sqrt{xy}$ or the product rule of radicals, the given expression, $ \sqrt[5]{16}\cdot\sqrt[5]{4} ,$ is equivalent to \begin{array}{l}\require{cancel} \sqrt[5]{16(4)} \\\\= \sqrt[5]{64} \\\\= \sqrt[5]{32\cdot2} \\\\= \sqrt[5]{(2)^5\cdot2} \\\\= 2\sqrt[5]{2} .\end{array}